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2 years ago

Solve the inequality. -1.5(4x+1) ≥ 45-25(x+1)

Mathematics

1 answer:

777dan777 [17]2 years ago

3 0

Answer:

x ≥ 21.5/19

Step-by-step explanation:

-1.5(4x + 1) ≥ 45 - 25(x + 1)

-6x - 1.5 ≥ 45 - 25x - 25

-6x + 25x ≥ 45 - 25 + 1.5

19x ≥ 21.5

x ≥ 21.5/19

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3 years ago

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3 years ago

Write each fraction as the sum of a whole number and a fraction less than 1

Tpy6a [65]

Given:

The fractions are:

$\dfrac{6}{5},\dfrac{11}{7},\dfrac{21}{4}$

To find:

The each fraction as the sum of a whole number and a fraction less than 1.

Solution:

The given fraction is $\dfrac{6}{5}$ .

$\dfrac{6}{5}=\dfrac{5+1}{5}$

$\dfrac{6}{5}=\dfrac{5}{5}+\dfrac{1}{5}$

$\dfrac{6}{5}=1+\dfrac{1}{5}$

Therefore, the given fraction $\dfrac{6}{5}$ can be written as $1+\dfrac{1}{5}$ .

The given fraction is $\dfrac{11}{7}$ .

$\dfrac{11}{7}=\dfrac{7+4}{7}$

$\dfrac{11}{7}=\dfrac{7}{7}+\dfrac{4}{7}$

$\dfrac{11}{7}=1+\dfrac{4}{7}$

Therefore, the given fraction $\dfrac{11}{7}$ can be written as $1+\dfrac{4}{7}$ .

The given fraction is $\dfrac{21}{4}$ .

$\dfrac{21}{4}=\dfrac{20+1}{4}$

$\dfrac{21}{4}=\dfrac{20}{4}+\dfrac{1}{4}$

$\dfrac{21}{4}=5+\dfrac{1}{4}$

Therefore, the given fraction $\dfrac{21}{4}$ can be written as $5+\dfrac{1}{4}$ .

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3 years ago